# Full text of "Treatise On Analysis Vol-Ii"

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```316       XV   NORMED ALGEBRAS AND SPECTRAL THEORY

An element x of A is of the form expO) for some y e A if and only if x is contained
in a connected commutative subgroup of G. (To prove that this condition is sufficient,
reduce to the case where A is commutative.)

10.    Let A be a Banach algebra with unit element e. Let G be the group of invertible
elements of A, and G0 the connected component of the neutral element of G.

(a)    If x e A is such that x" = e for some integer n > 0, show that x E G0. (Observe
that the spectrum of x is finite, and deduce that the set of complex numbers A such
that Xx + (1 — X)e is invertible is connected.)

(b)    Suppose that A is commutative. Show that every element of G/G0 (other than
the neutral element) has infinite order. (Remark that if x e G and x" e G0, then
x" = exp(j) for some ye A, whence (x Qxp(—(l/n)y))n = e; now use (a).)

11.    Let A be a commutative Banach algebra with unit element e. Let x e A and let
K = SpAC*:). Suppose that K has the following property: there exists a fundamental
system of neighborhoods Un of K such that the frontier of each Un is the union of a
certain number of disjoint simple closed curves which are images of circuits ynk
(1 ^ k ^Pn); the frontier of Un+1 is contained in Un, and if a function F is analytic
in a neighborhood of On — Un+i, with values in a complex Banach space, then

finally, if F is analytic in Un, then £ f   F(f) dt, - 0, and for all z e Un,

*=1 Jynk

F(z) = — V   f

27TI jft'l JVn..   .

(These properties can be proved by using Problem 4(b) of the Appendix to Chapter
IX; see Chapter XXIV for another proof.)

(a) Let V be a neighborhood of K, and/an analytic function on V with values in C.
Show that, for each n such that Un c V, the element

— £   f

2m * = i V

of A is independent of n. We denote it byf(x). If W is another neighborhood of K,
and g an analytic function in W with values in C, and if /(£) = g(£) for all £ in some
neighborhood of K contained in V n W, then f(x) = g(x).

(b)    If /(£) - 1, then /(*) = e. If /(£) = £, then f(x) == x. (Use Cauchy's theorem
(9.6.3) to reduce the problem to integration around a circle with center 0 and
radius >p(x), so as to be able to expand (£e — x)'1 as a power series in £~1.)

(c)   If h =/-f g (resp. h —fg}> where/and g are analytic in a neighborhood of K,
then h(x) = f(x)-+• g(x) (resp. h(x)=f(x)g(x)). (For the second assertion, express
f(x) in terms of a neighborhood Un and g(x) in terms of UnH. j; observe that

-i _     l

and use the Lebesgue-Fubini theorem.)

(d) If/is analytic in a neighborhood of K, with values in C, then Sp(/(#)) = /(Sp(x)).
(Observe that if A e Sp(x) we have/(A)e -f(x) = (Xe - x)g(x) for a suitably chosen
function g.) In particular, f(x) is invertible in A if and only if /(£) ^ 0 on Sp(#).n particular, the image of C under the
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